Anagrams string checker in Java

Carvia Tech | July 24, 2020 | 2 min read | 1,218 views | Java Coding Challenges


Anagram

Two strings are called anagrams if they contain same set of characters but in different order. For example,

  • peek and keep are anagrams

  • spar and rasp are anagrams

  • listen and silent are anagrams

Approach 1: sorting of characters

  1. Sort characters within string for both the inputs.

  2. if the two strings are same after sorting, they are anagram else not.

Let’s write Java code that takes this approach and checks if two given inputs are anagrams or not.

AnagramChecker.java
import java.util.Arrays;

public class AnagramChecker {

    public boolean isAnagram(String input1, String input2) {
        char[] sortedWord1 = input1.toCharArray();
        Arrays.sort(sortedWord1);

        char[] sortedWord2 = input2.toCharArray();
        Arrays.sort(sortedWord2);

        return Arrays.equals(sortedWord1, sortedWord2);
    }
}

Here is the unit test for above method.

JUnit testcase
import org.junit.Test;

import static org.hamcrest.CoreMatchers.equalTo;
import static org.junit.Assert.*;

public class AnagramCheckerTest {

    @Test
    public void isAnagram() {
        AnagramChecker checker = new AnagramChecker();

        assertThat(checker.isAnagram("peek", "keep"), equalTo(true));
        assertThat(checker.isAnagram("mary", "army"), equalTo(true));

        assertThat(checker.isAnagram("dart", "mart"), equalTo(false));
    }
}

Approach 2: Using count of characters for both words and compare

Alternatively, we can count the number of characters in both words and compare it for exactness. This approach is slightly efficient, as it has time complexity of Big O(n) compared to Big O(n log n) in earlier case.

AnagramEfficient.java
public class AnagramEfficient {

    public boolean isAnagram(String first, String second) {
        int[] letterCount = new int[126];

        for (char ch : first.toCharArray()) {
            letterCount[ch]++;
        }

        for (char ch : second.toCharArray()) {
            letterCount[ch]--;
        }

        for (int count : letterCount) {
            if (count != 0)
                return false;
        }
        return true;
    }
}

Here we are assuming that input is always an ASCII text without unicode support.

We can lowercase the input in both the approaches to make it case-insensitive.


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